Question

A nationwide survey of college students was conducted and found that students spend two hours per class hour studying. A professor at your university wants to determine whether the time students spend at your university is significantly different from the two hours. A random sample of 36 statistics students is carried out and the findings indicate an average of 1.75 hours. Assume a populatoin standard deviation of 0.5 hours. Using a level of significance of 0.05:

1. Construct a 95% confidence interval for the population mean.

2. Using a level of significance of 0.05: a. What is Ho and H1? b. What is the Decision Rule? c. What is the computed value of the test statistic? d. What is the decision regarding Ho? What can we conclude? e. Compute and Interpret the p-value.

3. Are the results from the confidence interval and hypothesis testing consistent? Should they be? Justify your answer.

Answer 1

1)

sample mean, xbar = 1.75
sample standard deviation, σ = 0.5
sample size, n = 36

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 0.5/sqrt(36)
ME = 0.1633

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (1.75 - 1.96 * 0.5/sqrt(36) , 1.75 + 1.96 * 0.5/sqrt(36))
CI = (1.59 , 1.91)

2)

a)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 2
Alternative Hypothesis, Ha: μ ≠ 2

b)

Rejection Region
This is two tailed test, for α = 0.05
Critical value of z are -1.96 and 1.96.
Hence reject H0 if z < -1.96 or z > 1.96

c)

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (1.75 - 2)/(0.5/sqrt(36))
z = -3

d)

Reject H0

e)
There is sufficient evidence to conclude that time students spend at your university is significantly different from the two hours.

f)

P-value Approach
P-value = 0.0027
As P-value < 0.05, reject the null hypothesis.

3)

yes, results are same because confidence interval does not contains 2